Motion of 3 Bodies Dynamics - Acclerations

[al_9591]

Dynamics. Motion of 3 bodies

Homework Statement

Initially the system of masses is held
motionless. All surfaces, and pulley are frictionless.
At the instant after the system of objects is released,
find the accelerations of m1, m2 and m3.
Data:
m1, m2 and m3

Homework Equations

F=ma

The Attempt at a Solution

EQUATIONS
FBD(free body diagram) of m1
X: Rcos(theta)=a1m1
Y: n1=n2+m1g
FBD of m2
X: T=(a2-a1)m2
Y: n2=m2g
FBD of m3
X: Tsin(theta)=a1
Y: Tcos(theta)-m3g=a2

This is what I have done so far are my equations correct?
Should I include the force n2 (normal 2) in the equation for the Y axis for mass 1?

 

[chaoseverlasting]

Your FBD for m3 is incorrect. The tension will not be at an angle theta. It'll be straight up.
Yes, you should include n2 in the Y axis equation for m1.

To find the acceleration of the block m1 you can use conservation of linear momentum.

Find acclerations of m2, m3; write out the equation for colm(m1,m2) and differentiate it wrt time. This gives you the relation between accelerations of m1 and m2, from which you can find out the acceleration of m1.

 

[zorro]

@chaoseverlasting
Can you explain how to find acceleration of mass m1 from conservation of linear momentum?

 

[chaoseverlasting]

As the surfaces are frictionless, the net force on the horizontal axis is zero. From the conservation of linear momentum you have, m1v1=m2v2.

Differentiating this equation you get m1a1=m2a2.

Now, you can find a2 from the second equation relating m3 and m2. Plugging that value into this equation gives you a1.

 

[zorro]

Oh Ic...thanks

 

[al_9591]

Hi, when you say second equation do you refer to the equation: T-m3g=a2?
Thank you for answering my question